Resonance — mechanical and electrical, conditions and applications

Question

What is resonance? Compare mechanical resonance (in SHM) with electrical resonance (in an LCR circuit). What is the condition for resonance in each case, and what happens to amplitude at resonance?

(JEE Main / NEET / CBSE Class 11 pattern)

Solution — Step by Step

Resonance occurs when a system is driven at its natural frequency — the amplitude of oscillation becomes maximum. Every oscillating system has a natural frequency; when the driving frequency matches it, energy transfer is most efficient.

flowchart TD
    A["Resonance"] --> B["Mechanical\n(spring-mass, pendulum)"]
    A --> C["Electrical\n(LCR circuit)"]
    B --> D["Condition:\nω_driving = ω₀ = √(k/m)"]
    C --> E["Condition:\nω_driving = ω₀ = 1/√(LC)"]
    D --> F["Amplitude maximum\n(limited by damping)"]
    E --> G["Current maximum\n(impedance minimum = R)"]
    F --> H["Analogy:\ndamping ↔ resistance\nmass ↔ inductance\nspring ↔ 1/capacitance"]

For a forced oscillator: $m\ddot{x} + b\dot{x} + kx = F_0\cos(\omega t)$

Natural frequency: $\omega_0 = \sqrt{k/m}$

At resonance ( $\omega = \omega_0$ ), the amplitude is:

A_{\text{max}} = \frac{F_0}{b\omega_0}

where $b$ is the damping coefficient. Less damping → sharper and taller resonance peak. With zero damping, amplitude would go to infinity (physically impossible — something would break).

For a series LCR circuit driven by AC voltage $V = V_0\sin(\omega t)$ :

Impedance: $Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2}$

At resonance, $\omega L = 1/\omega C$ , so:

\omega_0 = \frac{1}{\sqrt{LC}}

At this frequency: $Z = R$ (minimum impedance), current is maximum: $I_{\text{max}} = V_0/R$ .

The circuit behaves as if the inductor and capacitor cancel each other out, leaving only the resistance.

Why This Works

In both cases, resonance is about impedance matching. At the natural frequency, the reactive components (mass/spring in mechanics, L/C in circuits) cancel each other’s effects. Energy oscillates freely between the two storage modes (kinetic ↔ potential, or magnetic ↔ electric), and the driving force only needs to overcome dissipation (friction or resistance).

The mechanical-electrical analogy is precise: mass ↔ inductance (both resist change), spring constant ↔ 1/capacitance (both store potential energy), damping ↔ resistance (both dissipate energy). Understanding one system means you understand the other.

Alternative Method — Quality Factor

The sharpness of resonance is measured by the quality factor $Q$ :

Q = \frac{\omega_0}{\Delta\omega} = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}

Higher $Q$ means sharper resonance (narrow bandwidth). A radio tuner needs high $Q$ to select one station and reject nearby frequencies.

For JEE Main, the LCR resonance frequency $\omega_0 = 1/\sqrt{LC}$ is guaranteed to appear. Also remember: at resonance, the voltage across $L$ and $C$ individually can exceed the source voltage (voltage magnification), but they are equal and opposite, so they cancel. The voltage across $R$ equals the source voltage.

Common Mistake

Students often confuse the resonance frequency with the damped natural frequency. The natural frequency of an undamped system is $\omega_0 = \sqrt{k/m}$ . With damping, the natural frequency shifts slightly lower: $\omega_d = \sqrt{\omega_0^2 - (b/2m)^2}$ . For light damping, the difference is negligible, but for heavily damped systems, this matters. In LCR circuits, the resonance frequency remains exactly $1/\sqrt{LC}$ regardless of $R$ — resistance affects the peak height, not its position.

Question

Solution — Step by Step

Why This Works

Alternative Method — Quality Factor

Common Mistake

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